China Mathematical Olympiad

The minimum is $n+1$.

First, by taking $A=B=S$, we have $$|A\Delta S|+|B\Delta S|+|C\Delta S|=n+1.$$

Second, we can prove that $l=|A\Delta S|+|B\Delta S|+|C\Delta S|\ge n+1$. Let $X\setminus Y=\{x\mid x\in X,x\notin Y\}$. We have $$l=|A\setminus S|+|B\setminus S|+|C\setminus S|+|S\setminus A|+|S\setminus B|+|S\setminus C|.$$ All we need to prove are the following: (i) $|A\setminus S|+|B\setminus S|+|S\setminus C|\ge 1$, (ii) $|C\setminus S|+|S\setminus A|+|S\setminus B|\ge n$.

For (i). In fact, if $|A\setminus S|=|B\setminus S|=0$, then $A,B\subseteq S$. So $1$ cannot be an element of $C$, hence $|S\setminus C|\ge 1$, therefore (i) is valid.

For (ii). If $A\cap S=\varnothing$, then $|S\setminus A|\ge n$, the claim is already valid. If $A\cap S\ne \varnothing$, we assume that the maximal element of $A\cap S$ is $n-k$, $0\le k\le n-1$, then $$|S\setminus A|\ge k.\stackrel{◯}{1}$$ On the other hand, for $i=k+1,k+2,\dots ,n$, either $i\notin B$ (then $i\in S\setminus B$) or $i\in B$ (then $n-k+i\in C$, i.e., $n-k+i\in C\setminus S$), hence $$|C\setminus S|+|S\setminus B|⩾n-k.\stackrel{◯}{2}$$ From ① and ②, we obtain (ii).

In conclusion, (i) and (ii) are valid, so $l⩾n+1$. Hence, the minimum is $n+1$. $\square$